[ Note: the contents of this article overlap with a previous series on character theory. ]

Let K be a field and G a finite group. The group algebraK[G] is defined to be a vector space over K with basis , where “g” here is an abstract symbol for each element g of G. Thus, K[G] has dimension |G| over K. Now K[G] has a ring structure obtained from group multiplication, where the algebra product σ⋅τ is precisely the group product, i.e. an element of G. For example, multiplication in Q[S3] gives

The most important aspect of the group algebra is that its modules M correspond to group homomorphisms , of G to the group of invertible K-linear maps M→M; such a ρ is called a representation of G over K. Indeed, if M is a K[G]-module, then considering how elements g ∈ K[G] acts on M (for g ∈ G), we obtain a homomorphism

Conversely, if we have a group homomorphism , then this gives M the structure of a K[G]-module by letting () take m ∈ M to

Note that a homomorphism of K[G]-modules f : V → W is a linear map of vector spaces such that f(g⋅v) = g⋅f(v) for any g in G and v in V. Such a map is also called an intertwining map (since it commutes with every element of G).

Example 1

Every group has a trivial representation, corresponding to G → K* which maps all g to 1. As a K[G]-module representation, this is V := K and takes every v in V to

Example 2

V := K[G] is a module over the ring K[G]; this corresponds to the regular representation. E.g. if G = {e, g, g2} is the cyclic group of order 3, then the group homomorphism is given by:

E.g. g = (1,3,2) takes (x, y, z) to (y, z, x) and h = (1, 2) takes (x, y, z) to (y, x, z). The inverse of g is necessary for the subscripts to ensure that for any Now M is a module over R[S3], where, e.g. a(1, 2) + b(1, 3, 2) takes the point (x, y, z) to (ay+by, ax+bz, az+bx) so its corresponding matrix is assuming elements of M are written as column vectors.

Main Result

Theorem. If the characteristic of K does not divide |G|, then K[G] is semisimple.

Proof

We shall prove: for each left ideal I of K[G], there is a left ideal J such that I ⊕ J = K[G] (i.e. I ∩ J = 0 and I + J = K[G]). Assuming this holds, since K[G] is of finite dimension, it must have a minimal left ideal I1. By our assumption, there exists J1 such that I1 ⊕ J1 = K[G]. Again, J1 must have a simple submodule I2 etc . Eventually, we obtain K[G] as a direct sum of minimal left ideals.

Now suppose I is a left ideal of K[G]. This is a vector subspace, so we can write K[G] = I ⊕ W for a subspace W. Let p : K[G] → K[G] be the projection I ⊕ W → I, so that p2 = p and p has image I (see # later). Let us now define the K-linear map:

for

[ Note that this is well-defined since |G| is invertible in K. ] We have the following properties of q:

The image of q is in I : indeed image of p lies in I and g⋅I ⊆ I for any g in G.

q(α) = α for all α in I : indeed, and since p(α) = α for all α in I, the result follows. Together with property 1, this means im(q) = I.

q2 = q : for any α in K[G], we have q(α) in I, so by property 2, q(q(α)) = q(α).

q is a K[G]-module homomorphism: clearly q is K-linear; furthermore for any h in G, we have

Since q2 = q, q is a projection and we have K[G] = ker(q) ⊕ im(q) = ker(q) ⊕ I (see # later). And since q is a K[G]-module homomorphism, its kernel is a submodule, and we have proven our assumption. ♦

[ # Note: along the way, we used the fact that a K-linear map p : V → V such that p2 = p gives us V = ker(p) ⊕ im(p). The proof of this is an easy exercise. ]